微分積分学１ No.3 2004.10. 6
2. 関数のグラフと微分 2.1 微分係数と導関数（解答）
担当：市原問題 6 次の関数のx= 3における微分係数を, 定義に基づいてもとめなさい.
(1)y=x2+ 1
h→0lim
((3 +h)2+ 1)−(32+ 1)
h = lim
h→0
9 + 6h+h2+ 1−10
h = lim
h→0
6h+h2 h = lim
h→06 +h= 6
(2)y=√ x
h→0lim
√3 +h−√ 3
h = lim
h→0
(√
3 +h−√ 3)(√
3 +h+√ 3) h(√
3 +h+√
3) = lim
h→0
(3 +h)−3 h(√
3 +h+√ 3)
= lim
h→0
h h(√
3 +h+√
3) = lim
h→0
√ 1
3 +h+√ 3 = 1
2√ 3 =
√3 6
(3)y= 1 x
h→0lim 1 3 +h−1
3
h = lim
h→0
3−(3 +h) 3(3 +h)
h = lim
h→0
−h 3(3 +h)
h = lim
h→0
−h
3(3 +h)×(3(3 +h)) h×(3(3 +h))
= lim
h→0
−h
3h(3 +h)= lim
h→0
−1
3(3 +h) =−1 9
問題 7 次の関数を微分しなさい.
(1) y=x−4+x3 y0 =−4x−5+ 3x2
(2) y= 2√
x+x5 = 2x12 +x5 y0 = 2×1
2 ×x−12 + 5x4 =x−12 + 5x4
(3) y= (2√
x+x)(3x2+ 2x) = (2x12 +x)(3x2+ 2x)
= 6x12x2+ 4x12x+ 3x3+ 2x2 == 6x52 + 4x32 + 3x3+ 2x2
y0 = 15x32 + 6x12 + 9x2+ 4x
